Đặt VT bằng A
\(A^2=x-3+2\sqrt{\left(x-3\right)\left(5-x\right)}+5-x\)
\(A^2=2+2\sqrt{\left(x-3\right)\left(5-x\right)}\le2+\left(x-3\right)+\left(5-x\right)\)
\(A^2\le4\Leftrightarrow A\le2\)
Đặt VP=B
\(B=y^2+2.\sqrt{2013}.y+2013+2\)
\(B=\left(y+\sqrt{2013}\right)^2+2\ge2\)
mà A=B=2
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=5-x\\\left(y+\sqrt{2013}\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-\sqrt{2013}\end{matrix}\right.\)
\(\)